假设我有以下 XML 文档:
<species>
Mammals: <dog/> <cat/>
Reptiles: <snake/> <turtle/>
Birds: <seagull/> <owl/>
</species>
然后我得到这样的 species
元素:
import lxml.etree
doc = lxml.etree.fromstring(xml)
species = doc.xpath('/species')[0]
现在我想打印一份按物种分组的动物列表。我如何使用 ElementTree API 来做到这一点?
最佳答案
如果您枚举所有节点,您将看到一个带有类的文本节点,后跟带有物种的元素节点:
>>> for node in species.xpath("child::node()"):
... print type(node), node
...
<class 'lxml.etree._ElementStringResult'>
Mammals:
<type 'lxml.etree._Element'> <Element dog at 0xe0b3c0>
<class 'lxml.etree._ElementStringResult'>
<type 'lxml.etree._Element'> <Element cat at 0xe0b410>
<class 'lxml.etree._ElementStringResult'>
Reptiles:
<type 'lxml.etree._Element'> <Element snake at 0xe0b460>
<class 'lxml.etree._ElementStringResult'>
<type 'lxml.etree._Element'> <Element turtle at 0xe0b4b0>
<class 'lxml.etree._ElementStringResult'>
Birds:
<type 'lxml.etree._Element'> <Element seagull at 0xe0b500>
<class 'lxml.etree._ElementStringResult'>
<type 'lxml.etree._Element'> <Element owl at 0xe0b550>
<class 'lxml.etree._ElementStringResult'>
所以你可以从那里构建它:
my_species = {}
current_class = None
for node in species.xpath("child::node()"):
if isinstance(node, lxml.etree._ElementStringResult):
text = node.strip(' \n\t:')
if text:
current_class = my_species.setdefault(text, [])
elif isinstance(node, lxml.etree._Element):
if current_class is not None:
current_class.append(node.tag)
print my_species
结果
{'Mammals': ['dog', 'cat'], 'Reptiles': ['snake', 'turtle'], 'Birds': ['seagull', 'owl']}
这一切都很脆弱......文本节点排列方式的微小变化可能会扰乱解析。
关于python - 遍历 lxml etree 中的文本和元素,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24071072/